Physics, 6th Edition
Chapter 9 Impulse and Momentum
Chapter 9. Impulse and Momentum
9-1. A 0.5-kg wrench is dropped from a height of 10 m. What is its momentum just before it
strikes the floor? (First find the velocity from conservation of energy.)
mgh = ½mv2;
v  2 gh  2(9.8 m/s 2 )(10 m)
p = mv = (0.5 kg)(14 m/s);
v = 14.0 m/s
p = 7.00 kg m/s, down
9-2. Compute the momentum and kinetic energy of a 2400-lb car moving north at 55 mi/h.
m
W 2400 lb

; m = 75 slugs ; v = 55 mi/h = 80.7 ft/s
g 32 ft/s 2
p = mv = (75 slugs)(80.7 ft/s);
p = 6050 slug ft/s
K = ½mv2 = ½(75 slugs)(80.66 ft/s)2;
K = 244,000 ft lb
9-3. A 2500-kg truck traveling at 40 km/h strikes a brick wall and comes to a stop in 0.2 s. (a)
What is the change in momentum? (b) What is the impulse? (c) What is the average force
on the wall during the crash? Take + to be toward the wall. ( 40 km/h = 11.1 m/s)
p = mvf – mvo = 0 - (2500 kg)(11.1 m/s);
p = - 27,800 kg m/s
Impulse = p; F t = -27,800 kg m/s
Force ON truck:
F
27,800
;
0.2 s
Force on wall is opposite, so
F = -139,000 N
F = + 139,000 N
9-4. What is the momentum of a 3-kg bullet moving at 600 m/s in a direction 300 above the
600 m/s
horizontal? What are the horizontal and vertical components of this momentum?
300
p = 1800 kg m/s, 300
p = mv = (3 kg)(600 m/s);
px = 1800 cos 300 and py = 1800 sin 300;
101
px = 1560 kg m/s;
py = 900 kg m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-5. A 0.2-kg baseball traveling to the left at 20 m/s is driven in the opposite direction at 35
m/s when it is hit by a bat. The average force on the ball is 6400 N. How long was it in
contact with the bat?
35 m/s
( Impulse = change in momentum. )
+
F t = mvf – mvo = (0.2 kg)(35 m/s) – (0.2 kg)(-20 m/s)
(6400 N) t = 11 kg m/s;
t
t = 1.72 ms
-20 m/s
*9-6. A bat exerts an average force of 248 lb on 0.6-lb ball for 0.01 s. The incoming velocity of
the ball was 44 ft/s. If it leaves in the opposite direction what is its velocity?
Choose positive + as direction away from the bat, making incoming ball velocity negative:
F t = mvf – mvo ;
F t = mvf – mvo ;
0.6 lb
 0.01875 slugs
32 ft/s 2
m
(240 lb)(0.01 s) = (0.01875 slugs)vf - (0.01875 slugs)(-44 ft/s)
0.01875 vf = 2.4 lb s – 0.825;
vf = 84.0 ft/s
*9-7. A 500-g ball travels from left to right at 20 m/s. A bat drives the ball in the opposite
direction with a velocity of 36 m/s. The time of contact was 0.003 s. What was the
average force on the ball?
( m = 0.5 kg, vo = +20 m/s, vf = -36 m/s, t = 0.003 s )
F t = mvf – mvo ; F(0.003 s) = (0.5 kg)(-36 m/s) – (0.5 kg)(20 m/s)
F
18 kg m/s - 10 kg m/s
;
0.003 s
F = 9330 N
*9-8. A 400-g rubber ball is dropped a vertical distance of 12 m onto the pavement. It is in
contact with the pavement for 0.01 s and rebounds to a height of 10 m. What is the total
change in momentum? What average force is exerted on the ball?
To apply the impulse-momentum theorem, we need to first find the velocities
just before and just after impact with the ground.
102
Physics, 6th Edition
Chapter 9 Impulse and Momentum
(Ep)Beginning = (Ek)Ground ;
mgho = ½mvo2;
vf
12 m
v0  2 gh0  2(9.8 m/s2 )(12 m)
½mvf2 = mghf ;
Ft = mvf – mvo;
hf
vo = - 15.3 m/s
v f  2(9.8 m/s2 )(10 m)
10 m
vo
vf = + 14 m/s
F(0.01 s) = (0.4 kg)(14 m/s) – (0.4 kg)(-15.3 m/s);
F = 1170 N
*9-9. A cue stick strikes an eight-ball with an average force of 80 N over a time of 12 ms. If the
mass of the ball is 200 g, what will be its velocity?
Ft = mvf – mvo; (80 N)(0.012 s) = (0.2 kg)vf – 0;
v = 4.80 m/s
9-10. A golfer hits a 46 g golf ball with an initial velocity of 50 m/s at 300. What are the x- and
y-components of the momentum imparted to the ball?
300
vx = (50) cos 300 = 43.3 m/s ; vy = (50) sin 300 = 25.0 m/s
px = (0.046 kg)(43.3 m/s); py = (0.046 kg)(25 m/s)
50 m/s
vy
vx
px = 1.99 kg m/s; py = 1.15 m/s
*9-11. The face of the club in Problem 9-10 is in contact with the ball for 1.5 ms. What are the
horizontal and vertical components of the average force on the ball?
We need to treat horizontal and vertical impulses and momenta separately:
From previous problem: po = 0, pf = 1.99 kg m/s, pfy = 1.15 kg m/s
1.99 kg m/s
;
0.0015 s
Fx = 1330 N
1.15 kg m/s
;
0.0015 s
Fy = 767 N
Fx t = pfx – pox =1.99 kg m/s;
Fx 
Fx t = pfx – pox =1.15 kg m/s;
Fy 
103
Physics, 6th Edition
Chapter 9 Impulse and Momentum
Conservation of Momentum
9-12. A spring is tightly compressed between a 6-kg block and a 2–kg block and then tied with a
string. When the string breaks, the 2-kg block moves to the right with a velocity of 9
m/s. What is the velocity of the 6-kg block?
v1
Total momentum is zero before and after the event.
0 + 0 = m1v1 + m2v2 ;
v1 
 m2 v2 (2 kg)(9 m/s)

;
m1
(6 kg)
m1 = 6 kg
kg
m2 = 2 kg
v2
v1 = - 3.00 m/s
9-13. Two masses, one three times that of the other, are compressed against a spring and then
tied together on a frictionless surface as shown in Fig. 9-8. The connecting string
breaks and sends the smaller mass to the left with a velocity of 10 m/s. What was the
velocity of the larger mass?
Momentum zero before and after: 0 + 0 = m1v1 + m2v2
v2 
m1v1 m(10 m/s)

;
m2
(3m)
10 m/s m
v1
3m
v2
v1 = - 3.33 m/s
9-14. A 70-kg person standing on a frictionless surface throws a football forward with a velocity
of 12 m/s. If the person moves backward at 34 cm/s, what was the mass of the football?
Momentum zero before and after: 0 + 0 = m1v1 + m2v2
m2 
m1v1 (70 kg)(0.34 m/s)

;
v2
(-12 m/s)
m2 = 1.98 m/s
9-15. A 20-kg child is at rest in a wagon. The child jumps forward at 2 m/s, sending
the wagon backward at 12 m/s. What is the mass of the wagon?
0 = m1v1 + m2v2 ;
m1 
m2 v2 (20 kg)(2 m/s)

;
v1
(-12 m/s)
104
m1 = - 3.33 kg
Physics, 6th Edition
Chapter 9 Impulse and Momentum
9-16. Two children, weighing 80 and 50 lb, are at rest on roller skates. The larger child pushes
so that the smaller moves away at 6 mi/h. What is the velocity of the larger child?
0 = m1v1 + m2v2 ;
v1 
m2 v2 (50 lb)(6 ft/s)

;
m1
(80 lb)
v1 = - 3.75 ft/s
(Here were able to use the weight because it is proportional to the mass)
9-17. A 60-g firecracker explodes, sending a 45-g piece to the left and another to the right with a
velocity of 40 m/s. What is the velocity of the 45-g piece?
The two pieces add to 60 g: m1 + m2 = 60 g.
0 = m1v1 + m2v2 ;
v1 
Thus, m1 = 45 g, m2 = 15 g
m2 v2 (15 g)(40 m/s)

;
m1
(45 g)
v1 = - 13.3 m/s
*9-18. A 24-g bullet is fired with a muzzle velocity of 900 m/s from a 5-kg rifle. Find the recoil
velocity of the rifle and the ratio of the kinetic energy of the bullet to that of the rifle?
0 = m1v1 + m2v2 ;
v1 
m2 v2 (24 g)(900 m/s)

;
m1
(5000 g)
Ekb ½ mb vb2
(24 g)(900 m/s)2
;


Ekr ½ mr vr2 (5000 g)(4.32 m/s)2
v1 = - 4.32 m/s
Ratio = 208
*9-19. A 6-kg bowling ball collides head on with 1.8-kg pin. The pin moves forward at 3 m/s
and the ball slows to 1.6 m/s. What was the initial velocity of the bowling ball?
mbub + 0 = mbvb + mpvp; (6 kg)ub = (6 kg)(1.6 m/s) + (1.8 kg)(3 m/s)
6ub = 9.6 m/s + 5.4 m/s;
105
ub = 2.50 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-20. A 60-kg man on a lake of ice catches a 2-kg ball. The ball and man each move at 8 cm/s
after the ball is caught. What was the velocity of the ball before it was caught? What
energy was lost in the process? (A completely inelastic collision: vc = vm = vb = 8 cm/s)
mbub + mmum = (mb + mm)vc ; (2 kg)ub + 0 = (2 kg + 60 kg)(0.08 m/s)
2ub = 4.96 m/s;
ub = 2.48 m/s
½mbub2 + 0 =(mb + mm)vc2; ½(2 kg)(2.48 m/s)2 = ½(62 kg)(0.08 m/s)2 + Loss
Loss = 6.15 J – 0.198 J;
Loss = 5.95 J
*9-21. A 200-g rock traveling south at 10 m/s strikes a 3-kg block initially at rest. (a) If the two
stick together on collision, what will be their common velocity? (b) What energy was
lost in the collision?
mrur + mbub = (mr + mb)vc ; (0.2 kg)(10 m/s) + 0 = (0.2 kg + 3 kg)vc
2 m/s = 3.2 vc ;
vc = 0.625 m/s
½mrur2 + 0 =(mr + mb)vc2; ½(0.2 kg)(10 m/s)2 = ½ (3.2 kg)(0.625 m/s)2 + Loss
Loss =10.0 J – 0.625 J;
Loss = 9.38 J
Elastic and Inelastic Collisions
9-22. A car traveling at 8 m/s crashes into a car of identical mass stopped at a traffic light. What
is the velocity of the wreckage immediately after the crash, assuming the cars stick
together?
( u1 = 8.00 m/s; u2 = 0, m1 = m2 = m)
0
mu1 + mu2 = (m + m)vc ; mu1 = 2mvc
vc 
u1 8 m/s

;
2
2
106
vc = 4.00 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
9-23. A 2000-kg truck traveling at 10 m/s crashes into a 1200-kg car initially at rest. What is the
common velocity after the collision if they stick together? What is the loss in energy?
m1u1 + m2u2 = (m1 + m2)vc ; (2000 kg)(10 m/s) + 0 = (2000 kg + 1200 kg)vc
20,000 m/s = 3200 vc ;
vc = 6.25 m/s
½m1u12 + 0 =(m1 + m2)vc2; ½(2000 kg)(10 m/s)2 = ½(3200 kg)(6.25 m/s)2 + Loss
Loss = 100,000 J – 62,500 J;
Loss = 37,500 J
9-24. A 30-kg child stands on a frictionless surface. The father throws a 0.8-kg football with a
velocity of 15 m/s. What velocity will the child have after catching the football?
m1u1 + 0 = m1v1 + m2v2; (0.8 kg)(15 m/s) = (30 kg + 0.8 kg)vc
(30.8 kg)vc = 12 m/s;
vc =0.390 m/s
*9-25. A 20-g object traveling to the left at 8 m/s collides head on with a 10-g object traveling to
the right at 5 m/s. What is their combined velocity after impact?
m1u1 + m2u2 = (m1 + m2)vc ; (20 g)(-8 m/s) + (10 g)(5 m/s) = (20 g + 10 g)vc
-110 m/s = 30 vc ;
vc = -3.67 m/s, to left
*9-26. Find the percent loss of energy for the collision in Problem 9-25.
Conservation of Energy: ½m1u12 + ½m2u22 =½(m1 + m2)vc2 + Loss
½(20 g)(-8 m/s)2 + ½(10 g)(5 m/s)2 = ½(30 g)(-3.67 m/s)2 + Loss
765 J = 202 J + Loss;
Loss = 563 J;
107
%Loss =
563 J
= 73.6%
765 J
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-27. A 2-kg block of clay is tied to the end of a string as shown in
Fig. 9-9. A 500-g steel ball embeds itself into the clay causing both
to rise a height of 20 cm. Find the entrance velocity of the ball?
h
Before applying momentum conservation, we need to know the common velocity of the
clay and ball after the collision. Energy is conserved : ½(m1 + m2) vc2 = (m1 + m2) gh
vc  2 gh  2(9.8 m/s 2 )(0.20 m) ;
m1u1 + 0 = (m1 + m2) vc ;
vc = 1.98 m/s
(0.5 kg) u1 = (0.5 kg + 2 kg)(1.98 m/s)
(0.5 kg)u1 = 4.95 m/s;
u1 = 9.90 m/s
*9-28. In Problem 9-27, suppose the 500-g ball passes entirely through the clay an emerges with
a velocity of 10 m/s. what must be the new entrance velocity if the block is to raise to the
same height of 20 cm?
We must find the velocity v2 of the clay (m2) after collision:
2
10 m/s
½(m1 + m2) v2 = (m1 + m2) gh
vc  2 gh  2(9.8 m/s 2 )(0.20 m) ;
h
v2 = 1.98 m/s;
Momentum is conserved: m1u1 + 0 = m1v1 + m2v2;
(0.5 kg)u1 = (0.5 kg)(10 m/s) + (2 kg)(1.98 m/s);
u1 = 17.9 m/s
*9-29. A 9-g bullet is embedded into a 2.0 kg ballistic pendulum (see Fig. 8-13). What was the
initial velocity of the bullet if the combined masses rise to a height of 9 cm?
½(m1 + m2) vc2 = (m1 + m2) gh
vc  2 gh  2(9.8 m/s 2 )(0.09 m) ;
m1u1 + 0 = (m1 + m2) vc ;
vc = 1.33 m/s
(0.009 kg) u1 = (0.009 kg + 2 kg)(1.33 m/s)
(0.009 kg)u1 = 2.68 m/s;
108
u1 = 297 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-30. A billiard ball moving to the left at 30 cm/s collides head on with another ball moving to
the right at 20 cm/s. The masses of the balls are identical. If the collision is perfectly
elastic, what is the velocity of each ball after impact? (Consider right as +.)
Momentum: m1u1 + m2u2 = m1v1 + m2v2
Given: m1 = m2 = m, v1 = -30 cm/s, v2 = 0
m(-30 cm/s) + 0 = mv1 + mv2 ; v1 + v2 = (-30 cm/s) + (20 cm/s); v1 + v2 = -10 cm/s
Energy (e = 1): v2 – v1 = u1 – u2 = (-30 cm/s) – (20 cm/s); v2 – v1 = - 50 cm/s
From second equation: v2 = v1 – 50 cm/s; Substituting this for v2, we obtain:
v1 + (v1 - 50 cm/s) = - 10 cm/s;
and
And, v2 = v1 – 50 cm/s = (20 cm/s) – 50 cm/s;
v1 = 20 cm/s, to right
v2 = -30 cm/s, to left
9-31. The coefficient of restitution for steel is 0.90. If a steel ball is
h1
dropped from a height of 7 m, how high will it rebound?
e
h2
;
h1
e2 
h2
;
h1
h2  h1e2  (7 m)(0.9)2 ;
7m
h2
h2 =5.67 m
*9-32. What is the time between the first contact with the surface and the second contact for
Problem 9-31?
(We need to know vo to rise to 5.67 m, then find t.)
½ mv02  mgh; v0  2 gh  2(9.8 m/s 2 )(5.67 m) ;
s
v0  v f
2
t; t 
t = 1.07 s;
2s
2(5.67 m)

;
vo  0 10.54 m/s
T = 2t;
109
T = 2.15 s
vo = 10.54 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-33. A ball dropped from rest onto a fixed horizontal plate rebounds to a height that is 81
percent of its original height. What is the coefficient of restitution? What is the required
velocity on the first impact to cause the ball to rebound to a height of 8 m.
e
e
v2  v1
;
u1  u2
h2
 0.81;
h1
v2 = u2 = 0;
e = 0.900
e
v1
;
u1
u1 
v1
v1

;
e
(0.9)
½ mv12  mgh; v1  2 gh  2(9.8 m/s 2 )(8 m) ;
u1 = -1.11v1;
+
h1
u1 = -1.11(-12.5 m/s);
h2
u1 = -1.11v1
v1 = -12.5 m/s
u1 = 13.9 m/s
*9-34. A 300-g block moving north at 50 cm/s collides with a 200-g block moving south at 100
cm/s. If the collision is completely inelastic, what is their common velocity after sticking
together? What is the loss in energy? (Consider north as positive)
Momentum: m1u1 + m2u2 = m1v1 + m2v2 ;
v1 = v2 = vc for inelastic collision
(300 g)(50 cm/s) + (200 g)(-100 cm/s) = (300 g + 200 g)vc
15,000 g cm/s – 20,000 g cm/s = (500 g)vc;
vc = -10 cm/s, south
(Note: When working with energy, it is necessary to use kg for the mass unit.)
Conservation of Energy: ½m1u12 + ½m2u22 =½(m1 + m2)vc2 + Loss
½(0.3 kg)(-8 m/s)2 + ½(0.2 kg)(5 m/s)2 = ½(0.3 kg + 0.2 kg)(-3.67 m/s)2 + Loss
Solving for “loss”, we obtain:
Loss = 0.135 J
*9-35. Suppose the collision in Prob. 9-34 is perfectly elastic. Find the velocities after impact.
m1u1 + m2u2 = m1v1 + m2v2;
m1 = 300 g, m2 = 200 g, u1 = 50 cm/s, u2 = - 100 cm/s
(300 g)(50 cm/s) + (200 g)(-100 cm/s) = (300 g)v1 + (200 g)v2
Dividing each term by 100 g:
110
3 v1 + 2 v2 = -50 cm/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
Energy (e = 1): v2 – v1 = u1 – u2 = (50 cm/s) – (-100 cm/s); v2 – v1 = 150 cm/s
Substitute v2 = v1 + 150 cm/s into the earlier equation and solve for v1:
3 v1 + 2 (v1 + 150 cm/s) = - 50 cm/s;
v2 – (-80 cm/s) = 150 cm/s;
v1 = -80 cm/s, to left
v2 = 70 cm/s, to right
*9-36. A 5-lb and a 12-lb object approach each other with equal velocities of 25 ft/s. What will
be their velocities after impact if the collision is (a) completely inelastic or (b) perfectly
elastic? Since weight is proportional to mass, we will use the weights instead.
Momentum: W1u1 + W2u2 = W1v1 + W2v2 ;
v1 = v2 = vc for inelastic collision
(5 lb)(25 ft/s) + (12 lb)(-25 ft/s) = (5 lb + 12 lb)vc;
vc = -10.3 ft/s
Elastic case: (5 lb)(25 ft/s) + (12 lb)(-25 ft/s) = (5 lb)v1 + (12 lb)v2 ;
Dividing each term by 5 lb: v1 + 2.4 v2 = -35 ft/s
Energy (e = 1): v2 – v1 = u1 – u2 = (25 ft/s) – (–25 ft/s); v2 – v1 = 50 ft/s
Substitute v1 = v2 - 50 ft/s into the earlier equation and solve for v1:
(v2 – 50 ft/s) + 2.4 v2 = - 35 ft/s;
v1 = v2 – 50 ft/s = 4.41 ft/s – 50 ft/s;
v2 = 4.41 ft/s
v1 = -45.6 ft/s
Challenge Problems
9-37. An average force of 4000 N acts on a 400-g piece of metal causing it to move from rest to
a velocity of 30 m/s. What was the time of contact for this force?
F t = mvf – mvo = (0.4 kg)(30 m/s) – 0;
(4000 N)t = 12 kg m/s;
111
t = 3.00 ms
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-38. An 600-g object whose velocity is initially 12 m/s, collides with a wall and rebounds
with half of its original kinetic energy. What impulse was applied by the wall?
½mv02 = 2(½mvf2) ;
vf 
v02
(12 m/s)2

;
2
2
v f  8.49 m/s
F t = mvf – mvo = (0.6 kg)( –2.45 m/s) – (0.6 kg)(12 m/s);
F t = -12.3 N m
*9-39. A 10-kg block at rest on a horizontal surface is struck by a 20-g bullet moving at 200
m/s. The bullet passes entirely through the block and exits with a velocity of 10 m/s.
v2 = ?
What is the velocity of the block?
v1 = 10 m/s
m1u1 + 0 = m1v1 + m2v2; m1 = 0.02 kg
(0.02 kg)(200 m/s) = (0.02 kg)(10 m/s) + (10 kg)v2 ;
v2 = 0.380 m/s
9-40. In Problem 9-39, how much kinetic energy was lost?
Conservation of Energy: ½m1u12 + 0 = m1v12 + m2v22 + Loss
½(0.2 kg)(200 m/s)2 = ½(0.2 kg)(10 m/s)2 + ½(10 kg)(0.380 m/s)2 + Loss
Solving for “loss”, we obtain:
Loss =3990 J
*9-41. A 60-g body having an initial velocity of 100 cm/s to the right collides with a 150-g body
moving to the left at 30 cm/s. The coefficient of restitution is 0.8. What are the
velocities after impact. What percent of the energy is lost in collision?
m1u1 + m2u2 = m1v1 + m2v2;
m1 = 60 g, m2 = 150 g, u1 = 100 cm/s, u2 = - 30 cm/s
(60 g)(100 cm/s) + (150 g)(-30 cm/s) = (60 g)v1 + (150 g)v2
Divide each term by 60 g and simplify:
v1 + 2.5 v2 = 25 cm/s
v2 – v1 = e(u1 – u2); v2 – v1 = 0.8[100 cm/s – (-30 cm/s)]; v2 – v1 = 104 cm/s
Solve for v1:
v1 = v2 – 104 cm/s;
112
Now substitute to find v2.
Physics, 6th Edition
Chapter 9 Impulse and Momentum
(v2 – 104 cm/s) + 2.5 v2 = 25 cm/s;
*9-41. (Cont.)
v2 = 36.9 cm/s, to right
v1 = v2 – 104 cm/s = (36.9 cm/s) – 104 cm/s;
v1 = -67.1 cm/s, to left
Conservation of Energy: ½m1u12 + ½m2u22 = m1v12 + m2v22 + Loss
For energy we must use SI units with mass in “kg” and velocity in ”m/s.”
Eok = ½(0.06 kg)(1 m/s)2 + ½(0.15 kg)(-0.3 m/s)2;
Eok = 0.03675 J
Efk = ½(0.06 kg)(-0.671 m/s)2 + ½(0.15 kg)(0.369 m/s)2; Efk = 0.0237 J
 Eok  E fk
% Loss  100 
 Eok

 0.03675 J - 0.0237 J 
  100 

0.03675 J



%Loss = 35.5%
*9-42. The block in Fig. 9-6 weighs 6 lb. How high will it rise if a 0.4-lb projectile with an
initial velocity of 90 ft/s embeds itself into the block?
Momentum is conserved: m1u1 + 0 = m1v1 + m2v2;
(0.4 lb)(90 ft/s2) = (0.4 kg + 6 lb)vc
mW
h
vc = 5.625 ft/s
Now, we can find h using conservation of energy and the common initial velocity vc:
½(m1 + m2) v22 = (m1 + m2) gh ; The mass divides out.
h
v02 (5.625 ft/s) 2
;

2g
2(32 ft/s 2 )
h = 0.494 ft;
h = 5.93 in.
*9-43. A single railroad car moving north at 10 m/s strikes two identical, coupled cars initially
moving south at 2 m/s. If all three couple together after colliding, what is their common
velocity?
m1 = m2 = m3 = m;
u1 = 10 m/s;
u2 = u3 = -2 m/s;
v1 = v2 = v3 =vc
Momentum is conserved: mu1 + m(u2 + u3) = (3m)vc (Mass divides out.)
10 m/s – 2 m/s – 2 m/s = 3 vc;
113
vc = 2.00 m/s, north
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-44. An atomic particle of mass 2.00 x 10-27 kg moves with a velocity of 4.00 x 106 and
collides head on with a particle of mass 1.20 x 10-27 kg initially at rest. Assuming a
perfectly elastic collision, what is the velocity of the incident particle after the collision?
m1u1 + 0 = m1v1 + m2v2;
m1 = 2 x 10-27 kg, m2 = 1.2 x 10-27 kg, u1 = 4 x 106 m/s
(2 x 10-27 kg)( 4 x 106 m/s) = (2 x 10-27 kg)v1 + (1.2 x 10-27 kg )v2
Dividing each term by 2 x 10-27 kg:
v1 + 0.6 v2 = 4 x 106 m/s
Energy (e = 1): v2 – v1 = u1 – u2 = 4 x 106 m/s – 0 ; v2 – v1 = 4 x 106 m/s
Substitute v2 = ( v1 + 4 x 106 m/s) into the earlier equation and solve for v1:
v1 + 0.6 (v1 + 4 x 106 m/s) = 4 x 106 m/s;
v1 = 1.00 x 106 m/s
*9-45. A bat strikes a 400-g softball moving horizontally to the left at 20 m/s. It leaves the bat
with a velocity of 60 m/s at an angle of 300 with the horizontal. What are the horizontal
and vertical components of the impulse imparted to the ball?
First find components of velocity: v1x = - 20 m/s; v1y = 0
v2x = (60 cos 300) = 52.0 m/s; v2y = (60 sin 300) = 30 m/s
Fx t = J = mv2x – mv1x ; Jx = (0.4 kg)(52.0 m/s) – (0.4 kg)(-20 m/s);
Fy t = J = mv2y – mv1y ;
Jy = (0.4 kg)(30 m/s) – 0 ;
60 m/s
v2y
300
v1x
v2x
-20 m/s
Jx = 28.8 N s
Jy = 12.0 N s
*9-46. If the bat in Problem 9-45 was in contact with the ball for 5 ms, what was the magnitude
of the average force on the softball?
Fx 
28.8 N s
 5760 N ;
0.005 s
Fy 
12.0 N s
 2400 N
0.005 s
F  Fx2  Fy2  (5760 N)2  (2400 N)2 ;
114
F = 6240 N
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-47. A cart A has a mass of 300 g and moves on a frictionless air track at 1.4 m/s when it hits
a second cart B at rest. The collision is perfectly elastic and the 300-g cart’s velocity is
reduced to 0.620 m/s after the collision. What was the mass of the other cart and what
was its velocity after the collision?
m1 = 300 g;
u1 = 1.4 m/s; v1 = 0.620 m/s
m1u1 + 0 = m1v1 + m2v2; (300 g)(1.4 m/s) = (300 g)(0.620 m/s) + m2v2
m2v2 = 234 g m/s;
Elastic collision: v2 – v1 = u1 – u2 = (1.4 m/s) – 0
v2 = v1 + 1.4 m/s = 0.620 m/s + 1.4 m/s;
m2 v2  234 g m/s;
m2 
v2 = 2.02 m/s
234 g m/s
2.02 m/s
m2 = 116 g
*9-48. If the collision in Fig. 9-10, assume that the collision of the two masses is completely
inelastic. What is the common velocity after the collision and what is the ratio of the
15 m/s
final kinetic energy to the initial kinetic energy?
m1u1 + 0 = (m1 + m2)vc ; u1 = 15 m/s
(1 kg)(15 m/s) + 0 = (1 kg + 2 kg)vc
1 kg
2 kg
vc = 5.00 m/s
Ek 2 ½(m1  m2 )vc2
(3 kg)(5 m/s)2


;
Ek1
½ m1u12
(1 kg)(15 m/s)2
Ratio = 0.333
*9-49. Assume the collision in Problem 9-48 is perfectly elastic. What is the velocity of each
mass after the collision?
Elastic: m1u1 + 0 = m1v1 + m2v2 and v2 – v 1 = u1 – u2
(1 kg)(15 m/s) = (1 kg)v1 + (2 kg)v2 ;
v1 + 2v2 = 15 m/s ;
v1 = 15 m/s – 2 v2
v2 – v 1 = u1 – u2 = (15 m/s) – 0; v2 = 15 m/s + v1
v2 = 15 m/s + (15 m/s – 2v2);
v2 = 10 m/s ; v1 = 15 m/s – 2(10 m/s ) = -5 m/s
v1 = -5 m/s
and
115
v2 = 10 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-50. A 2-kg mass moves to right at 2 m/s and collides with a 6-kg mass moving to the left at 4
m/s. (a) If the collision is completely inelastic, what is their common velocity after
colliding, and how much energy is lost in the collision?
Momentum: m1u1 + m2u2 = m1v1 + m2v2 ;
v1 = v2 = vc for inelastic collision
(2 kg)(2 m/s) + (6 kg)(-4 m/s) = (2 kg + 6 kg)vc
4 kg m/s – 24 kg m/s = (8 kg)vc;
vc = -2.50 m/s
Conservation of Energy: ½m1u12 + ½m2u22 = ½(m1 + m2)vc2 + Loss
½(2 kg)(2 m/s)2 + ½(6 kg)(-4 m/s)2 = ½(2 kg + 6 kg)(-2.50 m/s)2 + Loss
Solving for “loss”, we obtain:
Loss = 27.0 J
**9-51. In Problem 9-50, assume the collision is perfectly elastic. What are the velocities after
the collision?
m1u1 + m2u2 = m1v1 + m2v2;
m1 = 2 kg, m2 = 6 kg, u1 = 2 m/s, u2 = - 4 m/s
(2 kg)(2 m/s) + (6 kg)(-4 m/s) = (2 kg)v1 + 6 kg)v2
Dividing each term by 2 kg:
v1 + 3 v2 = -10 m/s
Energy (e = 1): v2 – v1 = u1 – u2 = (2 m/s) – (-4 m/s); v2 – v1 = 6 m/s
Substitute v2 = v1 + 6 m/s into the earlier equation and solve for v1:
v1 + 3(v1 + 6 m/s) = - 10 m/s;
v2 – (-7.00 m/s) = 6 m/s;
116
v1 = -7.00 m/s
v2 = -1.00 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
Critical Thinking Questions
*9-52. An astronaut in orbit outside a capsule uses a revolver to control motion. The astronaut
with gear weighs 200 lb on the earth. If the revolver fires 0.05-lb bullets at 2700 ft/s, and
10 shots are fired, what is the final velocity of the astronaut? Compare the final kinetic
energy of the ten bullets with that of the astronaut. Why is the difference so great?
0 + 0 = Wava + Wbvb; va 
Wb vb (0.05 lb)(2700 ft/s)

; va = -0.675 ft/s
Wa
200 lb
Each shot changes va by –0.675 m/s: vf = 10(0.675 m/s);
We need masses: mb 
vf = - 6.75 ft/s
0.05 lb
200 lb
 0.00156 slugs ; mb 
 6.25 slugs
2
32 ft/s
32 ft/s 2
Ekb = 10 (½mvb2) = (5)(0.00156 slugs)(2700 ft/s)2; Ekb = 56,950 ft lb
Eka = ½mava2 = ½(6.25 slugs)(6.75 ft/s)2 ;
Eka = 142 ft lb
The kinetic energy of the bullets is much larger because when finding the kinetic
energy, one must deal with the square of velocity. The speeds dominate.
*9-53. In applying conservation of momentum for colliding objects to find final velocities, could
we use the weight of the objects instead of the mass? Why, or why not? Verify your
answer by applying it to one of the examples in the text.
Since weight is proportional to mass: W = mg, and since mass appears in every term
involving conservation of momentum, the weight can be used instead of the mass to
calculate either velocities or weights of colliding objects. For example, see Prob. 9-36.
m1u1 + m2u2 = m1v1 + m2v2
W1
W
W
W
u1  2 u2  1 v1  2 v2
g
g
g
g
W1u1 + W2u2 = W1v1 + W2v2
117
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-54. A 20-g bullet, moving at 200 m/s, strikes a 10-kg wooden block and passes entirely
through it, emerging with a velocity of 10 m/s. What is the velocity of the block after
impact?
How much energy was lost?
m1u1 + 0 = m1v1 + m2v2
(0.020 kg)(200 m/s) = (0.020 kg)(10 m/s) + (10 kg)v2 ;
v2 = 0.380 m/s
Conservation of Energy: ½m1u12 + 0 = ½(m1 + m2)vc2 + Loss
½(0.02 kg)(200 m/s)2 = ½(10.02 kg)(0.38 m/s)2 + Loss
Loss = 399 J
*9-55. A 0.30-kg baseball moving horizontally at 40 m/s is struck by a bat. The ball is in contact
with the bat for a time of 5 ms, and it leaves with a velocity of 60 m/s at an angle of 300,
what are the horizontal and vertical components of the average force acting on the bat?
First find components of velocity: v1x = - 20 m/s; v1y = 0
300
v2x = (60 cos 300) = 52.0 m/s; v2y = (60 sin 300) = 30 m/s
v1x
Fx t = mv2x – mv1x = (0.3 kg)(52.0 m/s) – (0.3 kg)(-40 m/s);
Fx(0.005 s) = 27.6 N s ;
60 m/s
v2y
v2x
-40 m/s
Fx = 5520 N
Fy t = mv2y – mv1y = (0.3 kg)(30 m/s) – 0 ; Fy(0.005 s) = 9.00 N s ;
Fy = 1800 N
*9-56. When two masses collide they produce equal but opposite impulses. The masses do not
change in the collision, so the change in momentum of one should be the negative of the
change for the other. Is this true whether the collision is elastic or inelastic. Verify your
answer by using the data in Problems 9-50 and 9-51.
Momentum is conserved whether energy is lost in collision or not. Therefore, equal but
opposite impulses should always produce equal and opposite changes in momentum.
Prob. 9-50: The test is whether: m1v1 – m1u1 = -( m2u2 – m2v2);
118
v1 = v2 = -2.5 m/s
Physics, 6th Edition
Chapter 9 Impulse and Momentum
*9-56. (Cont.)
(2 kg)(-2.50 m/s) - (2 kg)(2 m/s) = -[(6 kg )(-2.50 m/s) - (6 kg)(-4 m/s)]
- 9 kg m/s = - 9 kg m/s; It works for inelastic collisions.
Prob. 9-51: Same test: m1v1 – m1u1 = -( m2v2 – m2u2);
v1 = -7 m/s; v2 = -1 m/s
(2 kg)(-7 m/s) - (2 kg)(2 m/s) = -[(6 kg )(-1 m/s) - (6 kg)(-4 m/s)]
- 18 kg m/s = - 18 kg m/s; It also works for elastic collisions.
*9-57. Two toy cars of masses m and 3m approach each other, each traveling at 5 m/s. If they
couple together, what is their common speed afterward? What are the velocities of each
car if the collision is perfectly elastic? ( m1 = m, m2 = 3m, u1 = 5 m/s, u2 = - 5 m/s )
m1u1 + m2u2 = (m1 + m2 ) vc ;
-10 m/s = 4 vc;
m(5 m/s) + 3m(-5 m/s) = (m + 3m) vc
vc = -2.50 m/s
m1u1 + m2u2 = m1v1 + m2v2 ;
v1 + 3v2 = -10 m/s;
For inelastic case
m(5 m/s) + 3m(-5 m/s) = mv1 + 3mv2
Now for elastic: v2 – v1 = u1 – u2
v2 – v1 = 5 m/s – (–5 m/s) = 10 m/s;
v1 = v2 – 10 m/s
(v2 – 10 m/s) + 3 v2 = -10 m/s;
v1 = (0) – 10 m/s = - 10 m/s;
v2 = 0
v1 = -10 m/s
*9-58. An 8-g bullet is fired horizontally at two blocks resting on a frictionless surface. The first
block has a mass of 1-kg and the second has a mass of 2-kg. The bullet passes
completely through the first block and lodges into the second. After the collisions, the 1kg block moves with a velocity of 1 m/s and the 2-kg block moves with 2 m/s. What is
the velocity of the bullet before and after emerging from the first block?
--Observe the figure on the next page --
119
Physics, 6th Edition
Chapter 9 Impulse and Momentum
1 kg
8g
*9-58. (Cont.)
2 kg
2 m/s
1 m/s
Total momentum at start = total momentum at finish
(0.008 kg) v1 = (1 kg)(1 m/s) + (2.008 kg)(2 m/s);
v1 = 627 m/s
To find velocity emerging from 1-kg mass, we apply conservation to first block only:
(0.008 kg)(627 m/s) = (0.008 kg)ve + (1 kg)(1 m/s) ;
ve = 502 m/s
*9-59. A 1-kg mass A is attached to a support by a cord of length 80 cm, and it is held
horizontally as in Fig. 9-11. After release it swings downward striking the 2-kg mass B
which is at rest on a frictionless tabletop. Assuming that the collision is perfectly elastic
what are the velocities of each mass immediately after impact?
First find uA from energy of fall: ½mv2 = mgh
v  2 gh  2(9.8 m/s 2 )(0.8 m) ;
mAuA + 0 = mAvA + mBvB;
L = 80 cm
v = 3.96 m/s
mA = 2 kg
(1 kg)(3.96 m/s) = (1 kg)vA + (2 kg) vB
vA + 2 vB = 3.96 m/s ;
vB – vA = 3.96 m/s ;
mB = 2 kg
mA = 1 kg; uA = 3.96 m/s
Elastic: vB – vA = uA – uB = 3.96 m/s – 0
vA = vB – 3.96 m/s ;
Substitute for vA in the other equation.
(vB – 3.96 m/s) + 2 vB = 3.96 m/s; From which:
vA = vB – 3.96 m/s = 2.64 m/s – 3.96 m/s;
120
vB = 2.64 m/s
vA = -1.32 m/s